the number of left and right braces don't match..... please correct......
I have reached the following point with a statement negation:
Now, I understand that is the same as but can I move terms? For example in my negation, can I put the and the terms next to each other so that I get a true statement? or is there nothing more I can do (other than expanding with the )?
I couldn't find anything in my list of laws to answer this for me.
Negate the statement:
Sorry for confusion.
Just one thing. I can see how you have applied the distributive law to reduce the statement; however, one thing I don't see is how you know that the distribution of parentheses is like that.
As I complete my negation going through the steps I seem to arrive at:
I finish without any internal parentheses every time I do it :S
Strictly speaking, A \/ B /\ C is not a well-formed formula: it has to be either (A \/ B) /\ C or A \/ (B /\ C). We can drop parentheses only if we make a convention that, say, /\ binds stronger than \/; then A \/ B /\ C would stand for A \/ (B /\ C).
In this problem, the conjunction results from the negation of . So,
So for future instances when I come across problems that result in a conjunction without explicit usage of parentheses, like the above statement, I should assume parentheses around the terms always? I can see what you are saying for this problem, I am just trying to understand a general rule so when I come across something similar in the future, I will know what to do.
In all formal systems I can remember, A \/ B /\ C is interpreted as A \/ (B /\ C). However, this issue does not arise in this problem. Here, you don't start with . The (quantifier-free part of the) original formula is a conjunction of Qx and an implication. During negation, Qx and the implication are converted separately: Qx turns into ¬Qx, and the implication turns into a conjunction. Therefore, the negation is a disjunction of ¬Qx and this conjunction.
Then I change the implication to a conjunction first:
Then distribute the negative using De Morgan's:
and the parentheses disappear when I us De Morgans to leave an unclear conjunction if you see what I mean?
The parentheses do not disappear after the application of De Morgan's law. Remember that all subformulas have parentheses around them; some of them are just not written. So you take (¬(Qx)) \/ (¬((_) \/ (_))) and replace ¬((_) \/ (_)) with (_) /\ (_). The parentheses around ¬(_) are still there, so you get (¬(Qx)) \/ ((_) /\ (_)).
It is useful to think about formulas as tree structures. The strings of characters that we use to write formulas are just representations ("concrete syntax" in programming); what is important is the abstract structure. So you take the formula
and change the circled part to get
In the end, /\ is still performed first and the root \/ second.